Series solution for y"+x(e^-x) y = 0

**younhock** · April 5th 2010, 04:59 AM

Anyone can help me ?

Question: Find series solution about the origin for
y"+x(e^-x) y = 0
and determine the lower bound for the radius of convergence.

TQ.

**HallsofIvy** · April 5th 2010, 05:38 AM

Originally Posted by younhock

Anyone can help me ?

Question: Find series solution about the origin for
y"+x(e^-x) y = 0
and determine the lower bound for the radius of convergence.

TQ.

Presumably you know that $e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$ so that $xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$ .

Let $y= \sum_{n=0}^\infty a_nx^n$ so that $y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2}$ and put that into the differential equation.

**younhock** · April 5th 2010, 06:00 AM

Originally Posted by HallsofIvy

Presumably you know that $e^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!} x^n$ so that $xe^{-x}= \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$ .

Let $y= \sum_{n=0}^\infty a_nx^n$ so that $y"= \sum_{n=0}^\infty n(n-1)a_nx^{n-2}$ and put that into the differential equation.

But i do not know how to simplify the [ $\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{n+1}$ ] [ $\sum_{n=0}^\infty a_nx^n$ ] .

thanks for guiding...

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